On Friday, 1. February 2013 23:02:39 Ertugrul Söylemez wrote:

(f . g) x = f (g x)

so (f . g) x = f $ g x right? That looks like the two are pretty interchangeable. When would I prefer one over the other? -- Martin

Exactly. It is a convenience operator that lets you do: f x $ g y $ h z instead of f x (g y (h z)) As for whether you should write: foo = f . g or foo x = f $ g x ..go with whichever looks clearest and nicest to you. In most cases where you are just taking one argument and applying some functions to it, it's nice to omit the x. But in more complex cases, it may make your code harder to read and modify. See http://www.haskell.org/haskellwiki/Pointfree (and Problems with "pointless" style.) I personally always use pointfree where I would've required a lambda expression, e.g.: f (g . h) y instead of f (\x -> g $ h x) y On Tue, Feb 12, 2013 at 10:23 PM, Emanuel Koczwara < poczta@emanuelkoczwara.pl> wrote:

Hi,

Dnia 2013-02-12, wto o godzinie 22:09 +0100, Martin Drautzburg pisze:

...

On Friday, 1. February 2013 23:02:39 Ertugrul Söylemez wrote:

...

(f . g) x = f (g x)

so (f . g) x = f $ g x

right?

That looks like the two are pretty interchangeable. When would I prefer one over the other?

($) has lower precedence (it was introduced for that reason I belive).

Prelude> :info ($) ($) :: (a -> b) -> a -> b -- Defined in GHC.Base infixr 0 $

Please take a look at:

http://www.haskell.org/ghc/docs/latest/html/libraries/base-4.6.0.1/Prelude.h...

From the docs:

"Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted..."

Emanuel

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An example that helped me understand the utility of the $ operator is as follows: suppose that we have a list of functions fs :: [a -> b] fs = [f1, f2, ..., fk] and a list of arguments as :: [a] as = [a1, a2, ..., ak] and we want to produce the list [f1(a1), f2(a2), ..., fk(ak)], or equivalently [f1 a1, f2 a2, ..., fk ak], this can be done using the $ operator via zipWith ($) fs as simply because it makes the function application explicit. ::paul On 2013-02-12, at 13:44 , mukesh tiwari wrote:

You can write (f . g) x as f . g $ x so for me, it's avoiding extra parenthesis.

Mukesh

On Wed, Feb 13, 2013 at 2:53 AM, Emanuel Koczwara wrote: Hi,

Dnia 2013-02-12, wto o godzinie 22:09 +0100, Martin Drautzburg pisze:

...

On Friday, 1. February 2013 23:02:39 Ertugrul Söylemez wrote:

...

(f . g) x = f (g x)

so (f . g) x = f $ g x

right?

That looks like the two are pretty interchangeable. When would I prefer one over the other?

($) has lower precedence (it was introduced for that reason I belive).

Prelude> :info ($) ($) :: (a -> b) -> a -> b -- Defined in GHC.Base infixr 0 $

Please take a look at: http://www.haskell.org/ghc/docs/latest/html/libraries/base-4.6.0.1/Prelude.h...

From the docs:

"Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted..."

Emanuel

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A less obvious interpretation is to treat ($) as `id'. (f . g) x = f $ g x = f (id g x) = f (g x) Bah, we can even illustrate that ($) is simply an infix identity function: Prelude> map ((flip id) 2) [\x -> x - 3] [-1] Prelude> map (`id` 2) [\x -> x - 3] [-1] Prelude> map ($ 2) [\x -> x - 3] [-1] I just wanted to throw this out there as I found this out recently myself and found it fairly useful. I believe someone else already posted a link about pointless style. On 12/02/13 21:09, Martin Drautzburg wrote:

On Friday, 1. February 2013 23:02:39 Ertugrul Söylemez wrote:

...

(f . g) x = f (g x)

so (f . g) x = f $ g x

right?

That looks like the two are pretty interchangeable. When would I prefer one over the other?

Oops, you're right. Sorry for my oversight. On 13/02/13 08:12, Ertugrul Söylemez wrote:

Mateusz Kowalczyk wrote:

...

A less obvious interpretation is to treat ($) as `id'.

(f . g) x = f $ g x = f (id g x) = f (g x)

This is not how you get from ($) to id. The correct path is:

f $ x = ($) f x = f x = id f x = f `id` x

This equivalence is indicated by the type of ($). It's a specialized instance of a -> a:

($) :: (a -> b) -> (a -> b) ($) f = f

or equivalently:

($) :: (a -> b) -> a -> b ($) f x = f x

or equivalently:

($) :: (a ~ b -> c) => a -> a ($) = id

Greets, Ertugrul

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